A compound gave the following data of its composition by mass: C=57.82 %, O=38.58 % and the rest hydrogen. Its vapour density is 83. Empirical formula of compound is:
(C=12, O=16, H=1)

C_{6} H_{6} O_{4}

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C_{2} H_{4} O_{2}

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C_{4} H_{3} O_{2}

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C_{2} H_{3} O_{4}

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Solution

The correct option is C $C_{4} H_{3} O_{2}$
Molecular formula = 2 x V.D. = 2 x 83 = 166
% of Hydrogen $= 100 - (57.82 + 38.58) = 3.6$
Element % Ratio Simplest Ratio Empirical formula
Carbon $57.82 / 12 = 4.82$ $4.82 / 2.41 = 2$ $C_{4} H_{3} O_{2}$
Hydrogen $3.6 / 1 = 3.6$ $3.6 / 2.41 = 1.49$
Oxygen $38.58 / 16 = 2.41$ $2.41 / 2.41 = 1$
The empirical formula weight $= 12 \times 4 + 3 \times 1 + 16 \times 2 = 83$
Formula ratio $= 166 / 83 = 2$
The molecular formula of compound $= C_{8} H_{6} O_{4}$

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Similar questions

C = 57.28 %, O = 38.58 %

83

[C = 12, O = 16, H = 1]

An organisc compound, whose vapour density is 45, has the following percentage composition,
H = 2.22 % O = 71 . 19% and remaining carbon.

Calculate:
(a) its empirical formula, (b) its molecular formula.

C_{2} H_{4} O

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